$\begin{array}{1 1}(A)\;7500\\(B)\;7600\\(C)\;7700\\(D)\;7800\end{array} $

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- $n(C,r)=\large\frac{n!}{r!(n-r)!}$

Total number of men=10

Total number of women=7

Total number of members in the committee=6

The no of ways of at least 3 men and 2 women =$10C_3\times 7C_3+10C_4\times7C_2$

$10C_3=\large\frac{10!}{3!\times 7!}=\frac{10\times 9\times 8\times 7!}{7!\times 3\times 2}$

$\Rightarrow 120$

$7C_3=\large\frac{7!}{3!\times 4!}=\frac{7\times 6\times 5\times 4!}{3\times 2\times 4!}$

$\Rightarrow 35$

$10C_4=\large\frac{10!}{4!\times 6!}=\frac{10\times 9\times 8\times 7\times 6!}{4\times 3\times 2\times 6!}$

$\Rightarrow 210$

$7C_2=\large\frac{7!}{2!\times 5!}=\frac{7\times 6\times 5!}{ 2\times 5!}$

$\Rightarrow 21$

$10C_3\times 7C_3+10C_4\times 7C_2=12\times 35+210\times 21$

$\Rightarrow 4200+4410$

$\Rightarrow 8610$

The no of ways for 2 particular women to be always there =$10C_4+10C_3\times 5C_1$

$5C_1=\large\frac{5!}{1!4!}=$$5$

$10C_4+10C_3\times 5C_1=210+120\times 5$

$\Rightarrow 210+600$

$\Rightarrow 810$

The no of ways when two particular women are never together =total-together

$\Rightarrow 8610-810$

$\Rightarrow 7800$

Hence (D) is the correct answer.

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